Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long N = 998244353;
const double MAX = 1000000010;
double sy, a, b;
vector<double> starts;
vector<double> en;
vector<double> sums;
double res(double x, double y) {
  double res = 0;
  int n = starts.size();
  double l = a + (x - a) * (sy / (y + sy));
  double r = b + (x - b) * (sy / (y + sy));
  auto k = lower_bound(en.begin(), en.end(), l) - en.begin();
  if (starts[k] <= l) {
    res -= (l - starts[k]);
  }
  auto t = lower_bound(starts.begin(), starts.end(), r) - starts.begin();
  if (en[t - 1] > r) {
    res -= (en[t - 1] - r);
  }
  res += sums[t - 1] - sums[k - 1];
  res = res * (y + sy) / y;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> sy >> a >> b;
  sy = -sy;
  int n;
  cin >> n;
  int temp1, temp2;
  starts.push_back(-MAX);
  en.push_back(1 - MAX);
  for (int i = 0; i < n; i++) {
    cin >> temp1 >> temp2;
    starts.push_back(temp1);
    en.push_back(temp2);
  }
  starts.push_back(MAX);
  en.push_back(1 + MAX);
  sums.push_back(1);
  for (int i = 1; i < starts.size(); i++)
    sums.push_back(sums[i - 1] + en[i] - starts[i]);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> temp1 >> temp2;
    cout << setprecision(10) << res(temp1, temp2) << '\n';
  }
}